Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 40

Answer

$\Sigma_{n=1}^{\infty}(-1)^{n-1}\frac{x^{(3n+2)}}{n}$

Work Step by Step

$x^{2} \cdot ln(1+x^{3})=x^{2} \Sigma_{n=1}^{\infty}(-1)^{n-1}\frac{(x^{3})^{n})^{n}}{n}$ $=x^{2} \Sigma_{n=1}^{\infty}(-1)^{n-1}\frac{x^{3n}}{n}$ $=\Sigma_{n=1}^{\infty}(-1)^{n-1}\frac{x^{(3n+2)}}{n}$
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